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3 years ago

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A proton with an initial speed of 8.20×105 m/s is brought to rest by an electric field. Part APart complete Did the proton move

into a region of higher potential or lower potential? higher potential lower potential Previous Answers Correct Part BPart complete What was the potential difference that stopped the proton? Express your answer with the appropriate units.

1 answer:

Anastaziya [24]3 years ago

5 0

Answer:

(a) Region of higher potential

(b) ΔV=1.911V

Explanation:

For Part (a)

Proton moves into a region of higher potential because it must goes in opposite direction of force to get rest

For Part (b)

Potential difference we can calculate bu definition of kinetic energy

ΔK=mv²/2= -qΔV

express in ΔV

ΔV= -mv²/2q

Substitute the given values

So

ΔV= -mv²/2q

$=-\frac{mv^2}{2e}\\ =\frac{9.11*10^{-31}*(8.20*10^{5}m/s)^2}{2*1.602*10^{-19}}\\ =1.911V$

ΔV=1.911V

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A projectile is fired with a velocity of 22 m/s at an angle of 25°. What is the vertical component of the velocity?

7nadin3 [17]

Answer:

Vertical component of velocity is 9.29 m/s

Explanation:

Given that,

Velocity of projection of a projectile, v = 22 m/s

It is fired at an angle of 22°

The horizontal component of velocity is v cosθ

The vertical component of velocity is v sinθ

So, vertical component is given by :

$v_y=v\ sin(25)$

$v_y=22\ m/s\times\ sin(25)$

$v_y=9.29\ m/s$

Hence, the vertical component of the velocity is 9.29 m/s

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3 years ago

Constant Acceleration Kinematics: Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the dri

Ainat [17]

Answer:

The taken is $t_A = 19.0 \ s$

Explanation:

Frm the question we are told that

The speed of car A is $v_A = 22 \ m/s$

The speed of car B is $v_B = 29.0 \ m/s$

The distance of car B from A is $d = 300 \ m$

The acceleration of car A is $a_A = 2.40 \ m/s^2$

For A to overtake B

The distance traveled by car B = The distance traveled by car A - 300m

Now the this distance traveled by car B before it is overtaken by A is

$d = v_B * t_A$

Where $t_B$ is the time taken by car B

Now this can also be represented as using equation of motion as

$d = v_A t_A + \frac{1}{2}a_A t_A^2 - 300$

Now substituting values

$d = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

Equating the both d

$v_B * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

substituting values

$29 * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

$7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300$

$7 t_A =1.2 t_A^2 - 300$

$1.2 t_A^2 - 7 t_A - 300 = 0$

Solving this using quadratic formula we have that

$t_A = 19.0 \ s$

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3 years ago

Emmit is lifting a box vertically which forces are necessary for calculating the total force

zavuch27 [327]

Answer:

FG and FP

Explanation:

Gravitational Force(FG) because the box is being pulled down and a resistance force pushing up because he is pushing(FP) up on the box.

I hope this helps!

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3 years ago

Using the right amount of significant figures, calculate the answer to the following problem, 15.33879 + 15.555

MAXImum [283]

Answer:

there are 7 significant figures

Explanation:

15.33879+15.555

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there are 7 significant figures

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4 years ago

Determine e when I = 0.50 A and R = 12 W.

sammy [17]

Answer:

The correct answer is "24 V".

Explanation:

The given values are:

Current,

I = 0.50 A

Resistance,

R = 12 W

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On substituting the given values, we get

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3 years ago